Mathematics for junior high school with methods, and mathematics for junior high school with method

Isn't factoring very simple, why use the matching method?

Multiplication with crosses is simpler.

x^2-3)(3x^2-1)=0

If you want to use a matching method.

Divide by 3 on both sides at the same time

x^4-10/3x^2=-1

Recipe. x^4-10/3x^2+（10/6）^2=-1+（10/6）^2

x^2-10/6)^2=16/9

x^2-10/6=±4/3

x^2=10/6±4/3

x1=±（2√3）/3 x2=±√3/3

Matching method: let t = x 2

3t^2-10t +3=0

3(t^2- 10/3 t)+3=0

3(t^2-10/3t+25/9-25/9)+3=03(t^2-10/3t+25/9)-16/3=03(t-5/3)^2=16/3

t-5/3)^2=16/9

Namely. x^2-5/3)^2=16/9

x 2=1 3 or 3

So. x = root number 3 or negative root number 3 or root number 3 or negative root number 3.

Cross multiplication:

Let t=x 2

Then 3t 2 - 10t + 3 = 0

3t-1)(t-3)=0

Get. t=1 3 or t=3

i.e. x 2=1 3 or x 2=3

So x = root third or negative root third or root number three or negative root number three.

The formula for the third year of junior high school mathematics matching method = x + kx + n. The matching method refers to the transformation of a formula (including rational formula and transcendental travel formula) or a part of a formula into a perfect flat or several perfect flat forms through an identity deformation bridge. This method is often used in identity deformation to explore the implicit conditions in the problem, which is one of the powerful means to solve the problem.

In elementary algebra, collocation is a method used to reduce a quadratic polynomial to the sum of the squares of a primary polynomial and a constant. This method is to polynomialize the following forms into the coefficients a, b, c, d, and e in the above expression, which can also be expressions in their own right, and can be quickly controlled to contain variables other than x. The matching method is usually used to derive the formula for finding the root of a quadratic equation:

Our goal is to square the left side of the equation perfectly.

The formula for the third year of junior high school mathematics matching method = x + kx + n. The matching method refers to the transformation of a formula (including rational and transcendental) or a part of a formula into a perfect flat or several perfect flat sums through identity transformation. This method is often used in identity deformation to explore the implicit conditions in the problem, which is one of the powerful means to solve the problem.

In elementary algebra, the collocation method is a method used to reduce a quadratic polynomial to the sum of the squares of a primary multinomial grandband and a constant. This method is to polynomialize the following forms into the coefficients a, b, c, d, and e in the above expression, which can also be expressions themselves and can contain variables other than Zezi Lu x. The canonization method is usually used to derive the root finding formula for quadratic equations

Our goal is to square the left side of the equation perfectly.

The matching method is a solution method for solving a one-dimensional quadratic equation, that is, a one-dimensional quadratic equation is matched into a perfectly squared form, and then the square can be opened. For an equation where the quadratic term is 1, first move the constant term to the right side of the equation when formulating, then add half of the square of the primary term coefficient on both sides of the equation, and finally write the left side as perfectly squared, and solve the equation correctly.

Match perfectly squared, such as.

y=x-square+6x+8=(x-square+6x+9)-1=(x-square+2*3*x+3-square)-1=(x+3)square-1

y=4x+12x+18=[(2x) square+2*3*2x+3 square] +9 = (2x+3) square+9

And so on, can you understand? Let's take it.

Hello: Set the square side length x

x+3）（x+2）=2x²

x²+5x+6=2x²

x²-5x-6=0

x-6)(x+1)=0

x1 = 6 x2 = -1 (round).

Length 6+3=9cm

Width 6+2=8cm

Good luck with your studies!

Let the side length of the square be x

x+3)(x+2)=2x^2

x^2-5x-6=0

x1=6, x2=-1 (rounded).

The rectangle has a side length of 9,7

Solution: y2 2y=3

y2－2y＋4=3＋4

y－2）2=7

y 2 = plus or minus root number seven.

x1 root number seven 2

x2 negative root number seven 2

Simplifying the equation first yields y -2y + 3 = 0

y²-2y+1=2

y-1)²=2

y-1= root number 2

y-1= root number 2

Choose cy -2y+3=0, (y-1) =-2<0, no solution.

2(x-3 4) =-1+9 8=1 8>0, there is a solution, so x-3 4=+-1 4, so x=1 2 or 1

Answer: Unary quadratic equation: x px q=0

The matching method yields: x px= q

x²＋px＋﹙p／2﹚²=－q＋﹙p／2﹚²﹙x＋p／2﹚²=﹙p²－4q﹚／4

x＋p／2=±½√p²－4q﹚

x=－½p±½√p²－4q﹚

Where: p 4q 0

(a+b) = a +2ab+b is added twice as much as ab in the middle, and vice versa is half.

If it is an equation, the steps are as follows.

One. Organize the original equation into a general form: ax 2 + bx + c = 0 two. If the quadratic coefficient is not 1, divide both sides of the equation by a to make the quadratic coefficient 1 and three. Add half of the square of the primary term coefficient to both sides of the equation, so that the left side of the equation is perfectly squared.

Move the right side to the left side, and then formulate, get.

a-5) 2+(root(b-4)-1) 2+|(root number c-1)-2|=0

None of the terms can be less than 0, so it must be 0, then the equation is true and a=b=c=5 can be found